locus of orthocenters

**scipa** · August 11th 2008, 04:16 PM

Determine the locus of the point $O$ of intersection of the altitudes (orthocenter) of a triangle $ABC$ , if the locus of the vertex $A$ is a line parallel to $BC$ .

**earboth** · August 11th 2008, 10:18 PM

Originally Posted by scipa

Determine the locus of the point $O$ of intersection of the altitudes (orthocenter) of a triangle $ABC$ , if the locus of the vertex $A$ is a line parallel to $BC$ .

The locus of the orthocenter is a parabola passing through B and C.

Let $\alpha$ denote the interior angle at A, then

1. the parabola cuts the parallel to BC through A in 2 points if $\alpha > 90^\circ$ ;

2. the parallel to BC through A is tangent to the parabola if $\alpha = 90^\circ$ ;

3. the vertex of the parabola is located between the parallel to Bc and the line BC if $\alpha < 90^\circ$ ;

**earboth** · August 18th 2008, 01:27 AM

Originally Posted by scipa

Determine the locus of the point $O$ of intersection of the altitudes (orthocenter) of a triangle $ABC$ , if the locus of the vertex $A$ is a line parallel to $BC$ .

Second attempt:

Place the base BC of the triangle on the x-axis. Let d denote the distance between the 2 parallels. B(0,0) and C(c,0) and A(x,d).

The slope of AC is $m_{AC}=-\frac d{c-x}$ . Then the height through B (perpendicular to AC) has the slope:

$m_{h_B}=\frac{c-x}d$

Then the orthocenter is $H \left( x,\frac{(c-x)x}d\right)$

That means all orthocenters lie on the curve of

$y=-\frac1dx^2+\frac cdx$

I've choosen d = 5, C(8,0)

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